CLAMPERS

CLAMPERS
A clamper is a network constructed of a diode, a resistor and a
capacitor that shifts a waveform to a different dc level without
changing the appearance of the applied signal.
Analysis (ideal diode)
Operation at forward biased, the diode is short
circuited (i.e “on” state). The voltage will be
vo=0 since the current is shorted thru diode and
the capacitor is charged up to a voltage V.
Analysis
VC
Vdc
During reverse biased, the diode is open
circuited (i.e “off” state). The voltage will be
vo=0 since the current is shorted thru diode.
The voltage across R will be
Vdc + Vc= -V+(-V)=-2V
Result
Input
Output
Determine vo for the following network with the
input shown (for ideal diode).
Solution: Frequency is 1000Hz, then the period will be 1/f =
1ms ,so the interval for each level state is t1= 0.5ms. At first
interval the diode is open circuited, so no current at output,
therefore vo =0
Analysis (forward biased)
At 2nd interval, the diode is
will be the same as across
The voltage that charge up
-20V +Vc
short circuited, the voltage across R
the batery (parallel)
Vo= 5V
the capacitor, Applying KVL
-5V =0 , then VC=25V
The third interval will make the diode open circuited again and
current start to flow in the resistor (discharged the capacitor).
Applying the KVL
+10V +25V – vo=0
Give us
vo= 35V
Noted : the discharge time is can be determined as t= RC
RC=100kΩ x 0.1mF= 0.01s= 10ms
Total discharge 5t= 5x10ms=50ms which is >>interval time
which allow the capacitor to hold significantly the input
voltage.
The result
Practical case with diode of Vk=0.7V
At second interval vo = 5V-0.7V= 4.3V
and the charging up voltage -20V-5V +0.7V+Vc=0
Therefore
Vc= 24.3V
Circuit
result
The third interval we have
10V+24.3V-vo=0
Thus vo= 34.3V
Other example of clampers
The clamper also work well for sinusoidal wave.
ZENER DIODES
Showing the equivalent circuit at each state in V-I characteristic
Determine
(i) the voltages at references Vo1 and Vo2
(ii) the current thru LED and the power delivered by the supply
(iii) How does the power absorbed by the LED compare to that 6V
Zener diode
Vo1= VZ2 +VK= 3.3V +0.7V=4.0V
E=
Vo2=Vo1 +VK= 4V+ 6V= 10V
I R = I LED =
VR 40v − V02 − V LED 40v − 10V − 4V
=
=
= 20mA
R
1.3kΩ
1.3kΩ
Power delivered Ps=EIs=EIR= (40V)(20mA)=800mW
Absorbed by LED PLED=VLEDILED=(4V)(20mA)=80mW
Absorbed by Zener PZ=VZIZ= (6V)(20mA)=120mW
A LIMITER
Analysis
First half
2nd half
Fixed Vi and R as a dc regulator
A simplest Zener diode regulator network
To determine the state of Zener diode by removing the diode
from the network
Thus applying voltage divider rule
If V> VZ, the Zener diode is on.
If V< VZ, the Zener diode is off.
R LVi
V = VL =
R + RL
Zener equivalent for the “on” situation
IR =
V R Vi − V L
=
R
R
IL =
VL
RL
Since Zener is directly parallel to RL , then VL=VZ
Zener current , applying Kirchoff’s current law
Thus
And Power
IZ = IR – IL
PZ= VZ IZ
IR = IZ + IL
Ex: Determine VL , VR, IZ and PZ
Solution
Applying voltage divider rule V =
R LVi
1.2kΩ(16V )
=
= 8.73V
R + R L 1kΩ + 1.2kΩ
Since V=8.73V is less than 10V , the diode is in the “off” state
Thus VL=V=8.73V
And VR=Vi-VL=16V-8.73V=7.27V
Since the Zener is off , then IZ=0 and PZ= VZ IZ = 0W
Ex: Determine VL , VR, IZ and PZ
Applying voltage divider rule
R LVi
3kΩ(16V )
V =
=
= 12V
R + R L 1kΩ + 3kΩ
Since V=12V is greater than VZ=10V, the Zener is in “on” state
Therefore VL=VZ=10V
and VR= Vi- VL =16V -10V=6V
V L 10V
IL =
=
= 3.33mA
R L 3kΩ
and
VR
6V
IR =
=
= 6mA
R 1kΩ
I Z = I R − I L = 6mA − 3.33mA = 2.67 mA
To determine the resistor range
To determine the minimum load that can turn on the diode
So that VL=VZ ‘ that is
R LVi
V L = VZ =
RL + R
Solving for RL ‘ we have
R L min
RVZ
=
Vi − VZ
and
I L min
VL
VZ
=
=
R L R L min
Thus any resistance value greater than RLmin will ensure
that the Zener diode is in the “on” state
To determine the resistor range
Once the diode is in the “on” state, the voltage across R
remains fixed at
VR= Vi - VZ
And IR remains fixed at
The Zener current
VR
IR =
R
IZ = IR - IL
But the IZ is limited by the manufacturer IZM , then
ILmin = IR - IZM
And the maximum load resistance as
R L min =
VZ
I L min
Determine the range of RL and IL that will result
in VRL being maintained at 10V
Calculating for
minimum load
RLmin
R L min =
RVZ
(1kΩ )(10V ) = 10kΩ = 250Ω
=
40
Vi − VZ 50V − 10V
The voltage across the resistor R is VR= Vi – VZ =50V – 10V = 40V
This will give us
V R 40V
IR =
=
= 40mA
R 1kΩ
Continue
The minimum level of IL is ILmin = IR – IZM = 40mA -32mA = 8mA
Maximum load RLmax,
R L max =
VZ
I L min
10V
=
= 1.25kΩ
8mA
Power Pmax = VZ IZM= (10V )(32mA) = 320mW
Fixed RL and Variable Vi
The voltage Vi must be sufficiently large to turn the Zener diode
on. The minimum turn on voltage Vi= Vimin is
V L = VZ =
R LVi
RL + R
therefore
Vi min =
(RL + R )VZ
RL
Since the maximum Zener current IZM, Thus IZM=IR-IL
Then IRMAX = IZM + IL
The maximum voltage
Vi max = VR max + VZ
or
Vi max = I R max R + VZ
Determine the range of values of Vi that will maintain
the Zener diode of in the “on” state.
Using the formula given before
Vi min =
IL =
(RL + R )VZ
RL
(
1200Ω + 220Ω )(20V )
=
= 23.67V
1200Ω
V L VZ
20V
=
=
= 16.67 mA
R L R L 1.2kΩ
Vi max = I R max R + VZ
Continue
I R max = I ZM + I L = 60mA + 16.67 mA = 76.67 mA
Vi max = I R max R + VZ = (76.67 mA)(0.22kΩ ) + 20V = 36.87V
The Vi range is plotted below
If the input is a ripple from full-wave rectified and
filtering as shown, as long as within the specified
voltage, the output will still remain constant at 20V.
Voltage Multiplier
HALF-WAVE VOLTAGE DOUBLER
VC
VC
(a)During the positive voltage half-cycle across the transformer,
the diode D1 conducts and D2 is cut off. The capacitor C1 charge
up to peak rectified voltage Vm .
(b) Second half cycle, D2 conducts and D1 is cut-off. Now the
capacitor C2 is charged up with Vm + VC = Vm +Vm=2Vm
FULL-WAVE VOLTAGE DOUBLER
(a) Positive cycle, D1 is conducting, thus charging C1 to Vm . D2
is not conducting so charging on capacitor C2.
(b) Negative cycle, D2 is conducting, thus charging C2 to Vm.
D1 is not conducting so C1 still maintain the charging voltage
HALF-WAVE DOUBLER, TRIPLER AND QUADRUPLER
By arranging alternately capacitor and diode, we are able to
obtain voltage doubler, tripler and quadrupler. C1 plus
transformer charging C2. C2 charging C3 and C3 charging C4.
Protective configuration
Transient phase of a simple RL cct
Arcing during
opening the switch
Trying to change the current through an inductive element too
quickly may result in an inductive kick that could damage
surrounding elements or the system itself
The RL circuit may be used to control the relay
During closing the switch the coil will gain a steady current.
When closing, the arcing may cause the problem to the relay.
This is the cheapest circuit to protect the switching system.
Xc= 1/2πfC
Low cost ceramic
capacitor is usually used
A capacitor is parallel to the switch. It is acting as a
bypass ( or shorting) the high frequency component.
A snubber is also to short circuit the high frequency
component
The resistor in series is to protect the surge current.
Diode protection for RL circuit
A diode is placed parallel
to the inductive element
(relay). When switch
open the polarity of
voltage across coil will
turn on the diode thus
provide conduction path
for the inductor. The
diode must has the same
current level to that
current passing the coil
Diode protector to limit the emitter –base voltage
VBE is limited to 0.7V (knee voltage of the silicon diode)
Diode protection to prevent a reversal in collection current
A current from B to C will be blocked by the diode
Diodes can be used to limit the input of OPAM to 0.7V
Same appearance
Introduce voltage to increase limitation of the positive
portion and limit to 0.7V to the negative portion before
feeding to OPAM
Limit to 0.7V to negative portion
Limit to 6.7V to positive portion
POLARITY INSURANCE
This circuit is to prevent from mistaken connecting the battery
with wrong polarity
If the polarity is okay then the diode circuit is in open state
If the polarity is not okay then the current is bypass thru diode.
This will stop the battery to damage the $ system
Battery –powered backup
When electrical power is connected D1 id “on” state and D2
will be “off” state, thus only electrical power is functioned.
When electrical power is disconnected D1 is “off” state and D2
is conducted , thus the power will come from the battery.
Polarity detector using diodes and LED
For negative polarity red LED is lit
For positive polarity green LED is lit
LED diodes are arranged for EXIT sign display.
Voltage Reference Levels circuit
This circuit provide
different reference
levels
To establish a voltage level insensitive to the load current
A battery is connected to a network that has different
voltage supply and variable load. The battery available is
9V but the network require 6V. How?
Using external resistor
Let’s say the load is 1kΩ , then using voltage divider ,we determine
the value of external resistor we obtain approximately 470Ω
We calculate the VRL , give us 6.1V
Now if we change the load to 600 W but the external still same ,
then VRL become 4.9V … thus the system will not operate correctly!!
Using diode
Using diode the voltage can be converted using 4 silicon diode
which give a drop of voltage around 2.8V , thus the required
voltage of 6.2V is obtained. This network does not sensitive to the
load.
AC regulator and square-wave generator
Voltage across limit to 20V
conduct
Voltage across
corresponding to the
input if less than 20V
Same configuration to produce square -wave
Robert L. Boylestad
Electronic Devices and Circuit Theory, 9e
Copyright ©2006 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.